[3] Rigid Body Analysis

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1 [3] Rigid Body Analysis Page 1 of 53 [3] Rigid Body Analysis [3.1] Equilibrium of a Rigid Body [3.2] Equations of Equilibrium [3.3] Equilibrium in 3-D [3.4] Simple Trusses [3.5] The Method of Joints [3.6] Zero Force Members [3.7] The Method of Sections [3.8] Frames and Machines [3.9] Internal Loadings [3.10] Shear and Moment Diagrams

sum of the moments of the external forces at a point is zero: M 0 Q: How many scalar equations of equilibrium are available for 2-D Cartesian coordinates?

2 [3] Rigid Body Analysis Page 2 of 53 [3.1] Equilibrium of a Rigid Body From Newton s first law of motion, the necessary conditions of equilibrium can be defined as: a) The sum of the external forces acting on the body is zero: F 0 b) The sum of the moments of the external forces at a point is zero: M 0 Q: How many scalar equations of equilibrium are available for 2-D Cartesian coordinates? Q: How many scalar equations of equilibrium are available for 3-D Cartesian coordinates? O A: 3 and 6 (the number indicates the Degrees of Freedom, or DOF). Accurate free-body diagrams need to be constructed in order to solve the problems of a rigid body in equilibrium a) It must be isolated or cut free from its surroundings. b) It is necessary to include all the forces and couple moments applied on the body. c) It is required to replace supports with proper support reactions (unknown forces and moments). Note) A complete understanding of the free-body diagram is of PRIMARY importance for solving the problems in engineering mechanics. Various types of reactions occur at supports and points of support between bodies subjected to coplanar force systems: a) roller: one unknown reaction force which acts perpendicular to the contact surface b) pin: two unknowns, either two components or the magnitude and direction of the resultant force

[3] Rigid Body Analysis Page 3 of 53 c) fixed support: three unknowns, two components of the resultant force and a couple moment There are five important steps to draw a

b) Shaw all forces: all the external forces and couple moments (usually known) as well as all the support reactions (usually unknown) must be shown in the diagram.

3 [3] Rigid Body Analysis Page 3 of 53 c) fixed support: three unknowns, two components of the resultant force and a couple moment There are five important steps to draw a free-body diagram: a) Draw outlined shape: the body must be isolated or cut free from its supports and connections. b) Shaw all forces: all the external forces and couple moments (usually known) as well as all the support reactions (usually unknown) must be shown in the diagram. c) Identify external loadings: the forces and moments are all need to be labeled with their proper magnitudes and directions. d) Identify unknown reactions: the letters need to be assigned for all unknown forces and moments. e) Provide dimensions and coordinates.

4 [3] Rigid Body Analysis Page 4 of 53 CLASS EXAMPLE Draw the free-body diagram of member AB, which is supported by a roller at A and a pin at B. Explain the significance of each force on the diagram.

5 [3] Rigid Body Analysis Page 5 of 53 Name: Student ID: HOMEWORK Draw the free-body diagram of the beam which supports the 80 kg load and is supported by the pin at A and a cable which wraps around the pulley at D. Explain the significance of each force on the diagram.

[3] Rigid Body Analysis Page 6 of 53 CLASS EXAMPLE 3.1.

6 [3] Rigid Body Analysis Page 6 of 53 CLASS EXAMPLE Draw the free-body diagram of the bar, which has smooth points of contact at A, B, and C. Explain the significance of each force on the diagram.

7 [3] Rigid Body Analysis Page 7 of 53 Name: Student ID: HOMEWORK Draw the free-body diagram of member ABC which is supported by a smooth collar at A, roller at B, and short link CD. Explain the significance of each force acting on the diagram.

system), the scalar equations of equilibrium are given as: Fx 0, F y 0, and M O 0 The solution to some equilibrium problems can be simplified: a) Two-force

8 [3] Rigid Body Analysis Page 8 of 53 [3.2] Equations of Equilibrium In general, the equations of equilibrium for a rigid body can be defined as: F 0 and M O 0 In Cartesian 2-D coordinates (coplanar system), the scalar equations of equilibrium are given as: Fx 0, F y 0, and M O 0 The solution to some equilibrium problems can be simplified: a) Two-force members: a member is subject to no couple moments and forces are applied at only two points on a member: b) Three-force members: a member is subjected to only three forces, then it is necessary that the forces be either concurrent or parallel for the member to be in equilibrium:

9 [3] Rigid Body Analysis Page 9 of 53 CLASS EXAMPLE Determine the reactions at the supports.

10 [3] Rigid Body Analysis Page 10 of 53 Name: Student ID: HOMEWORK Determine the reactions at the supports.

11 [3] Rigid Body Analysis Page 11 of 53 CLASS EXAMPLE Determine the tension in the cable and the horizontal and vertical components of reaction of the pin A. The pulley at D is frictionless and the cylinder weights 80 lb.

12 [3] Rigid Body Analysis Page 12 of 53 Name: Student ID: HOMEWORK Determine the horizontal and vertical components of force at the pin A and the reaction at the rocker B of the curved beam.

13 [3] Rigid Body Analysis Page 13 of 53 CLASS EXAMPLE The horizontal beam is supported by springs at its ends. Each spring has a stiffness of k = 5 kn/m and is originally unstretched so that the beam is in the horizontal position. Determine the angle of tilt of the beam if a load of 800 N is applied at point C as shown.

14 [3] Rigid Body Analysis Page 14 of 53 Name: Student ID: HOMEWORK A man stands out at the end of the diving board, which is supported by two springs A and B, each having a stiffness of k = 15 kn/m. In the position shown the board is horizontal. If the man is 40 kg, determine the angle of tilt which the board makes with the horizontal. Neglect the weight of the board and assume it is rigid.

15 [3] Rigid Body Analysis Page 15 of 53 [3.3] Equilibrium in 3-D Various types of reactions occur at supports on a body subjected to 3-D force systems: In Cartesian 3-D coordinates, the scalar equations of equilibrium are given as: Fx 0, F y 0, and F z 0 M x 0, M y 0, and M z 0 In addition to satisfy the equations of equilibrium, the body must also be properly supported by its supports: a) Redundant supports: more supports than are necessary to hold the body in equilibrium (statically indeterminate). b) Improper supports: unknown forces are equal to the available equations of equilibrium, but instability can develop due to the improper supports

16 [3] Rigid Body Analysis Page 16 of 53 CLASS EXAMPLE Determine the x, y, z components of reaction acting on the ball-and-socket at A, the reaction at the roller B, and the tension in the chord CD required for equilibrium of the plate.

[3] Rigid Body Analysis Page 17 of 53 Name: Student ID: HOMEWORK 3.3.1 5-65 Determine the x, y, z components

17 [3] Rigid Body Analysis Page 17 of 53 Name: Student ID: HOMEWORK Determine the x, y, z components of reaction at the ball supports B and C and the ball-and-socket A (not shown) for the uniformly loaded plate.

[3] Rigid Body Analysis Page 18 of 53 CLASS EXAMPLE 3.3.2 Rod AB is supported by a ball-and-socket joint at A and a cable at B.

18 [3] Rigid Body Analysis Page 18 of 53 CLASS EXAMPLE Rod AB is supported by a ball-and-socket joint at A and a cable at B. Determine the x, y, z components of reaction at these supports if the rod is subjected to a 50-lb vertical force as shown.

19 [3] Rigid Body Analysis Page 19 of 53 Name: Student ID: HOMEWORK The boom is supported by a ball-and-socket joint at A and a guy wire at B. If the 5 kn loads lie in a plane which is parallel to the x-y plane, determine the x, y, z components of reaction at A and the tension in the cable at B.

planar trusses are often used to support roofs and bridges: There are two assumptions for the truss structure analysis: a) All loadings are applied at

20 [3] Rigid Body Analysis Page 20 of 53 [3.4] Simple Trusses A truss is a structure composed of slender members joined together at their end points: Gusset Plate Bolt or Pin Connection 2-D planar trusses are often used to support roofs and bridges: There are two assumptions for the truss structure analysis: a) All loadings are applied at the joints: if the member s weight is to be included in the analysis, it is represented by vertical forces (1/2 of weight applied at both ends). b) The members are joined together by smooth pins. Due to these two assumptions, each truss member acts as a twoforce member:

21 [3] Rigid Body Analysis Page 21 of 53 To prevent collapse, the form of a truss must be rigid: the simplest form that is rigid is a triangle

22 [3] Rigid Body Analysis Page 22 of 53 [3.5] The Method of Joints There are two methods of analysis available for the trusses: a) The method of joints: equilibrium of a joint of the truss, and the internal forces of the member becomes an external force on the joint b) The method of sections: equilibrium of the internal forces of the truss by cutting the members Because the truss members are all straight two-force members, the force system acting at each joint is coplanar and concurrent: Fx 0 and F y 0 Unknown internal forces developed in each member can be determined by the free-body diagram for each pin and equations of equilibrium: Joints Sections

23 [3] Rigid Body Analysis Page 23 of 53 [3.6] Zero Force Members Truss analysis using the method of joints is greatly simplified if zero-force members can be determined: a) The zero-force members support no loading. b) These zero-force members are used only to increase the stability of the truss during the construction and to provide support if the applied loading is changed. c) The zero-force members of a truss can generally be determined by inspection. There are two general rules to determine the zero-force members: a) If only two members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members

24 [3] Rigid Body Analysis Page 24 of 53 b) If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member (if no external force or support reaction is applied to the joint)

[3] Rigid Body Analysis Page 25 of 53 CLASS EXAMPLE 3.6.

25 [3] Rigid Body Analysis Page 25 of 53 CLASS EXAMPLE Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

26 [3] Rigid Body Analysis Page 26 of 53 Name: Student ID: HOMEWORK Determine the force in each member of the truss and state if the members are in tension or compression. Assume each joint as a pin. Set P = 4 kn.

27 [3] Rigid Body Analysis Page 27 of 53 CLASS EXAMPLE A sign is subjected to a wind loading that exerts horizontal forces of 300 lb on joints B and C of one of the side supporting trusses. Determine the force in each member of the truss and state if the members are in tension or compression.

28 [3] Rigid Body Analysis Page 28 of 53 Name: Student ID: HOMEWORK Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

29 [3] Rigid Body Analysis Page 29 of 53 [3.7] The Method of Sections The method of sections is used to determine the internal loadings acting within the member: a) If the body is in equilibrium, then any part of the body is in equilibrium b) The member of a truss can be cut and new free-body diagram is defined for that part

30 [3] Rigid Body Analysis Page 30 of 53 CLASS EXAMPLE Determine the force in members CD, CF, and FG of the Warren truss. Indicate if the members are in tension or compression.

[3] Rigid Body Analysis Page 31 of 53 Name: Student ID: HOMEWORK 3.7.

31 [3] Rigid Body Analysis Page 31 of 53 Name: Student ID: HOMEWORK The truss supports the vertical loading shown. Determine the force in members KJ, CD, and KD, and state if the members are in tension or compression.

32 [3] Rigid Body Analysis Page 32 of 53 [3.8] Frames and Machines Frames and machines are two common types of structures which are often composed of multiforce members: a) Frames are generally stationary and are used to support loads. b) Machines contain moving parts and are designed to transmit and alter the effect of forces. In order to determine the forces acting on a frame or a machine, the members need to be disassembled into parts and the free-body diagrams must be drawn for each parts: The equations of equilibrium are applied to the free-body diagram in order to determine unknown forces and moments at the supports.

33 [3] Rigid Body Analysis Page 33 of 53 CLASS EXAMPLE Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

34 [3] Rigid Body Analysis Page 34 of 53 Name: Student ID: HOMEWORK The compound beam is supported by a rocker at B and is fixed to the wall at A. If it is hinged (pinned) together at C, determine the reactions at the supports.

35 [3] Rigid Body Analysis Page 35 of 53 CLASS EXAMPLE Determine the horizontal and vertical components of force that the pins at A, B, and C exert on the frame. The cylinder has a mass of 80 kg.

[3] Rigid Body Analysis Page 36 of 53 Name: Student ID: HOMEWORK 3.8.2 6-101 The derrick is used to lift the 300 kg stone with constant velocity.

36 [3] Rigid Body Analysis Page 36 of 53 Name: Student ID: HOMEWORK The derrick is used to lift the 300 kg stone with constant velocity. If the derrick and the block and tackle are in the position shown, determine the horizontal and vertical components of force at the pin support A and the orientation and tension in the guy cable BC.

[3] Rigid Body Analysis Page 37 of 53 [3.

using free-body diagrams and equations of equilibrium: The design of structural or mechanical members

loadings can be determined by using the method of sections: a) If a cut is made for a 2-D member,

37 [3] Rigid Body Analysis Page 37 of 53 [3.9] Internal Loadings In the previous chapters, the equilibrium of rigid bodies has been analyzed using free-body diagrams and equations of equilibrium: The design of structural or mechanical members requires an analysis of the loading acting within the member (internal loading): The internal loadings can be determined by using the method of sections: a) If a cut is made for a 2-D member, three unknown forces and moments need to be applied at the cross section: b) If a cut is made a 3-D member, six unknown forces and moments need to be applied at the cross section:

38 [3] Rigid Body Analysis Page 38 of 53 It is important to keep in mind that you can only handle up to certain number of unknowns for each free-body diagram: a) 2-D problem: three b) 3-D problem: six

39 [3] Rigid Body Analysis Page 39 of 53 CLASS EXAMPLE Determine the internal normal force, shear force, and the bending moment in the beam at points C and D. Assume the support at B is a roller. Point C is located just to the right of the 8-kip load.

40 [3] Rigid Body Analysis Page 40 of 53 Name: Student ID: HOMEWORK Determine the internal normal force, shear force, and moment at point C.

41 [3] Rigid Body Analysis Page 41 of 53 CLASS EXAMPLE Determine the internal normal force, shear force, and the moment at points C and D.

42 [3] Rigid Body Analysis Page 42 of 53 Name: Student ID: HOMEWORK Determine the internal normal force, shear force, and moment at point C of the beam.

[3] Rigid Body Analysis Page 43 of 53 [3.

knowledge of the variation of the internal shear force V and bending moment M.

c) The normal forces are ignored in shear and moment diagrams.

43 [3] Rigid Body Analysis Page 43 of 53 [3.10] Shear and Moment Diagrams Beams are structural members which are designed to support loadings applied perpendicular to their axes: a) The design of beams requires a detailed knowledge of the variation of the internal shear force V and bending moment M. b) The variations of V and M as functions of the position x along the beam s axis can be obtained by the method of sections; shear and moment diagrams. c) The normal forces are ignored in shear and moment diagrams. d) Plotting the shear and moment diagrams requires to establish a sign convention: e) Shear and moment as functions of the position x can be determined by cutting the beam at the location x and applying the equations of equilibrium on that cut free-body diagram. Considering a beam under the arbitrary loadings, the relations between distributed load, shear, and moment can be mathematically derived. a) A small element x at location x is cut out as a free-body diagram. b) Applying equations of equilibrium on this free-body diagram yields: F y 0 : V w( x) x ( V V ) 0 V w( x) x

44 [3] Rigid Body Analysis Page 44 of 53 : V x M w( x) x[ k( x)] ( M M ) 0 O + M 0 M V x w( x) k( x) 2 Dividing by x and taking the limit as x 0, these two equations become: dv wx ( ) dx (Slope of shear diagram = Distributed load magnitude) dm V ( x ) dx (Slope of moment diagram = Shear magnitude) These equations can be integrated as: V w( x) dx (Change in shear = Area under the curve of distributed load) M V ( x) dx (Change in moment = Area under the curve of shear diagram) A free-body diagram of a small element of a beam, taken at one of the forces provides: F y 0 : V F Therefore, the shear will jump downward when force F acts downward on the beam.

45 [3] Rigid Body Analysis Page 45 of 53 A free-body diagram of a small element, taken at the couple moment provides: + M O 0 : M MO Therefore, the moment will jump upward when clockwise moment MO acts on the beam.

46 [3] Rigid Body Analysis Page 46 of 53 CLASS EXAMPLE Draw the shear and moment diagrams for the beam.

[3] Rigid Body Analysis Page 47 of 53 Name: Student ID: HOMEWORK 3.10.

47 [3] Rigid Body Analysis Page 47 of 53 Name: Student ID: HOMEWORK The suspender bar supports the 600 lb engine. Draw the shear and moment diagrams for the bar.

[3] Rigid Body Analysis Page 48 of 53 CLASS EXAMPLE 3.

48 [3] Rigid Body Analysis Page 48 of 53 CLASS EXAMPLE Draw the shear and moment diagrams for the beam.

49 [3] Rigid Body Analysis Page 49 of 53 Name: Student ID: HOMEWORK Draw the shear and moment diagrams for the beam.

[3] Rigid Body Analysis Page 50 of 53 CLASS EXAMPLE 3.

50 [3] Rigid Body Analysis Page 50 of 53 CLASS EXAMPLE Draw the shear and moment diagrams for the beam.

51 [3] Rigid Body Analysis Page 51 of 53 Name: Student ID: HOMEWORK Draw the shear and moment diagrams for the beam.

52 [3] Rigid Body Analysis Page 52 of 53 CLASS EXAMPLE The beam consists of two segments pin connected at B. Draw the shear and moment diagrams for the beam.

53 [3] Rigid Body Analysis Page 53 of 53 Name: Student ID: HOMEWORK Draw the shear and moment diagrams for the beam.

Beams. Lesson Objectives:

Beams. Lesson Objectives: Beams Lesson Objectives: 1) Derive the member local stiffness values for two-dimensional beam members. 2) Assemble the local stiffness matrix into global coordinates. 3) Assemble the structural stiffness